Can you give an example of a field extension $k\subseteq K$ such that, every element of $K$ is transcendental over k and $K$ is not separable over $k$?
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2$\begingroup$ Not a research question -- did you read the first question of the faq before asking? $\endgroup$– Julien PuydtCommented May 26, 2013 at 12:02
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$\begingroup$ If you think that's not a research question, then you are free to vote to close, but first please give me reference for the answer. $\endgroup$– anonymousCommented May 26, 2013 at 12:27
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2$\begingroup$ Let $k$ of char. $p>0$ have $[k:k^p] > p$; $a, b$ in a $p$-basis. Let $f=x^p+ay^p+b z^p$. Over $\overline{k}$, $f$ is the $p$th power of $x+a^{1/p}y+b^{1/p}z$, so $A=k[x,y,z]/(f)$ is a domain, normal by Serre's criterion; ${\rm{Spec}}(A)$ is geometrically irreducible, nowhere smooth over $k$. Thus, $K={\rm{Frac}}(A)$ isn't separable over $k$, $k$ is separably closed in $K$. To show $K\cap k^{1/p}=k$, if not we get $c\in k - k^p$ with $k(c^{1/p})\subset K$, so $k(c^{1/p})\subset A$ (normality!). Then $A\otimes_k k(c^{1/p})$ is non-reduced, so $a^{1/p},b^{1/p}\in k(c^{1/p})$, contradiction. $\endgroup$– user29283Commented May 26, 2013 at 15:53
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